133 lines
3.3 KiB
Markdown
133 lines
3.3 KiB
Markdown
# Lecture 28
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## Chapter VI Inner Product Spaces
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### Orthonormal basis 6B
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Example:
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Find a polynomial $q\in \mathscr{P}_2(\mathbb{R})$ such that
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$$
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\int^1_{-1}p(t)cos(\pi t)dt=\int^1_{-1}p(t)q(t)dt
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$$
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for $p\in \mathscr{P}_2(\mathbb{R})$
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note that $\varphi(p)=\int^1_{-1}p(t)cos(\pi t)cos(\pi t)dt$ is a linear functional. Thus by **Riesz Representation Theorem**, $\exists$ unique $q$ such that $\varphi (p)=\langle p,q \rangle=\int^1_{-1}pq$
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$$
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q=\overline{\varphi(e_0)}e_0+\overline{\varphi(e_z)}e_z
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$$
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where $e_0,e_1,e_z$ is an orthonormal basis.
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and $q=\frac{15}{2\pi^2}(1-3x^2)$
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#### Orthogonal Projection and Minimization
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#### Definition 6.46
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If $U$ is a subset of $V$, then the **orthogonal complement** of $U$ denoted $U^\perp$
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$$
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U^\perp=\{v\in V\vert \langle u,v\rangle =0,\forall u\in U\}
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$$
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The set of vectors orthogonal to every vector in $U$.
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#### Theorem 6.48
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Let $U$ be a subset of $V$.
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(a) $U^\perp$ is a subspace of $V$.
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(b) $\{0\}^\perp=V$
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(c) $V^\perp =\{0\}$
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(d) $U\cap U^\perp\subseteq\{0\}$
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(e) If $G,H$ subsets of $V$ with $G\subseteq H$, then $H^\perp\subseteq G^\perp$
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Example:
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Two perpendicular line in 2D plane.
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Let $e_1,...,e_m$ be an orthonormal list, let $u=Span(e_1,...,e_m)$ How do I find $U^\perp$?
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Extend to an orthonormal basis $e_1,...,e_m,f_1...,f_n$. $U^\perp=Span(f_1,...,f_n)$
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#### Theorem 6.40
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Suppose $U$ is finite dimensional subspace of $V$, then $V=U\oplus U^\perp$
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Proof:
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Note $U\cap U^\perp=\{0\}$, so it suffices to show $U+U^\perp=V$. Fix an orthonormal basis $e_1,...,e_m$ of $U$. Let $v\in V$, let $u=\langle v,e_1\rangle e_1+...+\langle v,e_m\rangle e_m$ let $w=v-u$, then $v=u+W$ we need to check that $w\in U^\perp$
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$\langle w,e_k \rangle=\langle v,e_k\rangle-\langle u,e_k\rangle=\langle v,e_k\rangle-\langle v,e_k\rangle=0$
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So $w\in U^\perp$
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#### Corollary 6.51
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$$
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dim\ U^\perp=dim\ V-dim\ U
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$$
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#### Theorem 6.52
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Let $U$ be a finite dimensional of a vector space $V$. Then $(U^\perp)^\perp=U$
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Proof:
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First let $u\in U$ we want to show $u\in (U^\perp)^\perp$, then $\langle u,w \rangle=0$ for all $w\in U^\perp$ but then $u\in (U^\perp)^\perp$
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Exxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxercise on the other directiononoooonononon.
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#### Corollary 6.54
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$U^\vert=\{0\}\iff U=V$
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Proof:
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$(U^\perp)^\perp=\{0\}^\perp\implies U=V$
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#### Definition 6.55
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Given $U$ a finite dimensional subspace of $V$. The **orthogonal projection of $V$ onto $U$** is the operator $P_u\in \mathscr{L}(V)$ defined by: For each $v$ write $v=u+w$ where $u\in U$ and $w\in U^\perp$ then $P_u v=u$
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Formula:
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Let $e_1,...,e_n$ an orthonormal basis of $U$.
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$P_u v=\langle v,e_1\rangle e_1+...+\langle v,e_m\rangle e_m$
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#### Theorem 6.57
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(a) $P_u$ is linear.
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(b) $P_u u=U,\forall u\in U$
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(c) $P_u w=0,\forall w\in U^\perp$
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(d) $range\ P_u=U$
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(e) $null\ P_u=U^\vert$
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(f) $v-P_u v\in U^\perp$
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(g) $P_u^2=P_u$
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(h) $||P_u v||\leq ||v||$
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Proof:
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(a) Let $v,v'\in V$ and suppose $v=u+w,v'=u'+w'$, then $v+u'=(u+u')+(w+w')$ this implies that $P_u(v+v')=u+u'=P_u v+ P_u v'$
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...
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#### Theorem 6.58 Riesz Representation Theorem
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Let $V$ be a finite dimensional vector space for $v\in V$ define $\varphi_v\in V'$ by $\varphi_v(u)=\langle u,v \rangle$. Then the map $v\to \varphi_v$ is a bijection.
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Proof:
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Surjectivity Ideal is let $w\in (null\ \varphi)^\perp$
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$$
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v=\frac{\varphi(w)}{||w||^2}w,\varphi(v)=||v||^2
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$$
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make sense $\varphi_v=\varphi$
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